Search Results

If $A$ is a $\mathbb{Q}$-algebra, then there is a group $G$ such that $A$ is a quotient of $\mathbb{Q}[G]$ if and only if $A$ is generated by units. For the "if" direction, take $G$ to be the group of …

If $\Lambda$ is a ring, then the isomorphism classes of finitely generated projective $\Lambda$-modules form a commutative monoid $(A,+)$, with $[P]+[Q]=[P\oplus Q]$. This monoid contains a distinguis …

The rings satisfying your condition (for right modules) are the right pure semisimple rings. There are many equivalent conditions. You can find a lot of information in Section 4.5 of the book Prest, M …

There are abelian groups $A$ such that $A\cong A\oplus A \oplus A$ but $A\not\cong A\oplus A$. Let $E=\operatorname{End}(A)$. The functor $F=\operatorname{Hom}(A,-)$ is an equivalence from the catego …

No. Let $R$ be the ring of eventually constant sequences of integers, and let $e_i\in R$ be the sequence whose $i^{th}$ term is 1, with all other terms zero. Then if $I$ is the annihilator of an eleme …

Every simple module is trivially quasi-projective, and if every simple $R$-module is projective then $R$ is semisimple. So semisimple rings are the only rings for which quasi-projective implies projec …

If $\Gamma$ acts $2$-transitively on an infinite set $X$, then the permutation module $\mathbb{Q}[X]$ will be a counterexample. For example, take an action of the free group of rank $2$ on a countabl …

$R$ doesn't need to be connected, so long as $k$ is (and if $R$ is connected then $k$ is, since a nontrivial idempotent of $k$ would be a nontrivial central idempotent of $R$). Also, $R$ doesn't need …

Let $k$ be a field, let $R$ be the path algebra over $k$ of the quiver $$1\xrightarrow{\gamma}2\xrightarrow{\delta}3$$ modulo the relation $\gamma\delta=0$, and let $e=e_2$ be the idempotent associate …

Since $S$ has infinite projective dimension, there is some indecomposable summand $M$ of $\Omega^n(S)$ that has infinite projective dimension. For simple modules $T$ of finite injective dimension, $\ …

There's a four-dimensional counterexample over any field. $A$ has basis $\{e,a,b,c\}$, with all products of basis elements zero except for $$e^2=e,\quad ab=c,\quad ea=a,\quad ec=c,\quad be=b,\quad ce= …

No, even if $S$ is commutative. There may be easier counterexamples, but ... There are commutative Noetherian local (and therefore semiperfect) rings $S$ with a finitely generated indecomposable modul …

This isn't an area that I'm expert on, and it's quite possible there's a much more elementary and/or more general answer. But if the Bass Conjecture on Hattori-Stallings ranks for group rings is true …

Even for a $2$-dimensional vector space (so we're looking at subrings of the ring $M_2(\mathbb{F})$ of $2\times2$ matrices over $\mathbb{F}$) there are nonisomorphic maximal commutative subrings. Bo …

I haven't thought about base changes, but the original problem for $\alpha=1$ (so $\sigma$ is the identity map and $R=\mathbb{Z}_p[[t]][F]$ is just a polynomial ring over $\mathbb{Z}_p[[t]]$, and clas …